In Young's double-slit experiment `d//D = 10^(-4)` (d = distance between slits, D = distance of screen from the slits) At point P on the screen, resulting intensity is equal to the intensity due to the individual slit `I_(0)`. Then, the distance of point P from the central maximum is `(lambda = 6000 Å)`

Options

A. 2 nm
B. 1 mm
C. 0.5 mm
D. 4 mm

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We are given:
- \( \frac{d}{D} = 10^{-4} \) (where \( d \) is the distance between the slits and \( D \) is the distance from the slits to the screen)
- \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \)
- The intensity at point \( P \) on the screen is equal to the intensity due to an individual slit, \( I_0 \).

### Step 2: Use the intensity relation
In Young's double-slit experiment, the resultant intensity \( I \) at a point on the screen is given by:
\[ I = I_{max} \cos^2\left(\frac{\phi}{2}\right) \]
where \( \phi \) is the phase difference.

Since \( I = I_0 \), we can write:
\[ I_0 = I_{max} \cos^2\left(\frac{\phi}{2}\right) \]

### Step 3: Find \( I_{max} \)
The maximum intensity \( I_{max} \) when both slits are open is given by:
\[ I_{max} = I_1 + I_2 + 2\sqrt{I_1 I_2} \]
Since \( I_1 = I_2 = I_0 \):
\[ I_{max} = I_0 + I_0 + 2\sqrt{I_0 I_0} = 4I_0 \]

### Step 4: Substitute \( I_{max} \) into the intensity equation
Now substituting \( I_{max} \) back into the intensity equation:
\[ I_0 = 4I_0 \cos^2\left(\frac{\phi}{2}\right) \]
Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ 1 = 4 \cos^2\left(\frac{\phi}{2}\right) \]
This simplifies to:
\[ \cos^2\left(\frac{\phi}{2}\right) = \frac{1}{4} \]

### Step 5: Find the phase difference \( \phi \)
Taking the square root:
\[ \cos\left(\frac{\phi}{2}\right) = \frac{1}{2} \]
This implies:
\[ \frac{\phi}{2} = \frac{\pi}{3} \quad \text{or} \quad \frac{\phi}{2} = \frac{5\pi}{3} \]
Thus:
\[ \phi = \frac{2\pi}{3} \quad \text{or} \quad \phi = \frac{10\pi}{3} \]

### Step 6: Relate phase difference to path difference
The phase difference \( \phi \) is related to the path difference \( \Delta x \) by:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting \( \phi = \frac{2\pi}{3} \):
\[ \frac{2\pi}{3} = \frac{2\pi}{\lambda} \Delta x \]
This simplifies to:
\[ \Delta x = \frac{\lambda}{3} \]

### Step 7: Relate path difference to distance on the screen
The path difference \( \Delta x \) can also be expressed as:
\[ \Delta x = \frac{y d}{D} \]
Setting the two expressions for \( \Delta x \) equal:
\[ \frac{y d}{D} = \frac{\lambda}{3} \]

### Step 8: Solve for \( y \)
Rearranging gives:
\[ y = \frac{\lambda D}{3d} \]
Substituting \( \frac{d}{D} = 10^{-4} \) implies \( D = 10^4 d \):
\[ y = \frac{\lambda (10^4 d)}{3d} = \frac{10^4 \lambda}{3} \]
Substituting \( \lambda = 6000 \times 10^{-10} \):
\[ y = \frac{10^4 \times 6000 \times 10^{-10}}{3} \]
Calculating:
\[ y = \frac{60000000 \times 10^{-10}}{3} = 20000000 \times 10^{-10} = 2 \times 10^{-3} \, \text{m} = 2 \, \text{mm} \]

### Final Answer:
The distance of point \( P \) from the central maximum is \( \mathbf{2 \, mm} \).

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